91Ó°ÊÓ

Hydrogen atoms ground state is described by${\mathbf{ψ}}_{\mathbf{100}}\left(r,\mathrm{θ},\mathrm{Ï•}\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{Ï€²¹}}^{\mathbf{3}}}}{\mathbf{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}$

(a)

The probability of finding the electron inside a nucleus with radius is:

$$\begin{gathered} P=∫{\overline{)\mathrm{ψ}}}^2{d}^{3}r \\ =\frac{1}{\mathrm{π}{a}^3}{∫}_0^{2\mathrm{π}}{∫}_0^{\mathrm{π}}{∫}_0^b{e}^{-2rla{r}^2}\sin \left(\mathrm{θ}\right)drd\mathrm{θ}d\mathrm{ϕ} \\ =\frac{4\mathrm{π}}{\mathrm{π}{a}^3}{∫}_0^b{e}^{-2rla{r}^2}dr \\ ={\overline{)\frac{4}{a^3}\left[-\frac{a}{2}{r}^2{e}^{-2rla}+\frac{a^3}{4}{e}^{-2rla}\left(-\frac{2r}{a}-1\right)\right]}}_0^b \\ =-{\overline{)\left(1+\frac{2r}{a}+\frac{2r^2}{a^2}\right)e^{-2rla}}}_0^b \\ =1-\left(1+\frac{2b}{a}+2\frac{2b^2}{a^2}\right)e^{-2rla} \\ P=1-\left(1+\frac{2b}{a}+2\frac{2b^2}{a^2}\right)e^{-2rla} \end{gathered}$$

Thus,

the probability is$1-\left(1+\frac{2b}{a}+2\frac{2b^2}{a^2}\right)e^{-2rla}$.

(b)

Assume $\mathrm{Î\mu }=2b/a$in the result of part (a), and get,

$$\begin{gathered} P=1-\left(1+∈+\frac{1}{2}{∈}^2\right)e^{-∈} \\ P=1-\left(1+∈+\frac{1}{2}{∈}^2\right)\left(1-∈\frac{{∈}^2}{2}-\frac{{∈}^3}{3!}\right) \\ P=1-\left(1+∈+\frac{1}{2}{∈}^2\right)\left(1-∈\frac{{∈}^2}{2}-\frac{{∈}^3}{3!}\right) \\ =1-1+∈-\frac{∈2}{2}+\frac{{∈}^7}{6}-∈+{∈}^2-\frac{∈2}{2}-\frac{∈2}{2}+\frac{∈2}{2} \\ ={∈}^3\left(\frac{1}{6}-\frac{1}{2}+\frac{1}{1}\right) \\ P=\frac{4}{3}{\left(\frac{b}{a}\right)}^3 \end{gathered}$$

Thus, the lowest order term is cubic.

(c)

Alternatively, let the wave function is constant across the nucleus' volume.

$$\begin{gathered} a≈0.5×{10}^{-10}m \\ =\frac{4}{3}\frac{\mathrm{π}{b}^3}{\mathrm{π}{a}^3} \\ =\frac{4}{3}{\left(\frac{b}{a}\right)}^3 \\ P=\frac{4}{3}{\left(\frac{b}{a}\right)}^{\mathbf{3}} \end{gathered}$$

Therefore, the result matches with the result of part b.

(d)

Calculate the probability by using$b≈{10}^{-15}$and$a≈0.5×{10}^{-10}$

$$\begin{gathered} P=\frac{4}{3}{\left(\frac{{10}^{-15}}{0.5×{10}^{-10}}\right)}^3 \\ =\frac{4}{3}{\left(2×{10}^{-5}\right)}^3 \\ =\frac{4}{3}.8×{10}^{-15} \\ =\frac{32}{3}×{10}^{-15} \\ P1.07×{10}^{-14} \end{gathered}$$

Hence, the estimate for P is$1.07×{10}^{-14}$.